Answer:
x = ± 3, x = ± 2i
Step-by-step explanation:
Given
f(x) = (x² + 4)(x² - 9)
To find the zeros let f(x) = 0, that is
(x² + 4)(x² - 9) = 0
Equate each factor to zero and solve for x
x² - 9 = 0 ( add 9 to both sides )
x² = 9 ( take the square root of both sides )
x = ± [tex]\sqrt{9}[/tex] = ± 3
x² + 4 = 0 ( subtract 4 from both sides )
x² = - 4 ( take the square root of both sides )
x = ± [tex]\sqrt{-4}[/tex] = ± [tex]\sqrt{4(-1)}[/tex] = ± [tex]\sqrt{4}[/tex] ×[tex]\sqrt{-1}[/tex] = ± 2i
Thus zeros are
x = - 3, x = 3 ← real
x = - 2i, x = 2i ← complex
